Basler Electric BE1-87B - Manual (Page 33)

The excitation currents, (I)1, (I)2,……(I)n will be a function of the peak voltages that can be produced in
the secondary of the respective CTs. It is possible to determine the currents with the aid of the secondary
excitation characteristic for the respective CT. But it is first necessary to modify the characteristics so that
they are plotted as a function of the peak voltages that can be produced. The procedure for doing so is
provided in the following paragraphs.
1. Determine the knee point coordinates of the standard excitation curve (E
S
and Ie). These points will
be indicated on the given characteristic, or they can be found graphically by determining the point
where a 45-degree line is tangent to the knee of the excitation curve.
2. Calculate and plot the following point on the same sheet with the excitation curve:
V = (7) (E
S
)
I = (5) (Ie) (EQUATION 7)
3. Draw a line having a slope of ½ through the point (V, I) calculated and plotted in step 2. A slope of ½
corresponds to one log cycle on the vertical axis (voltage) and two log cycles on the horizontal axis
(current) (See line A-B in Figure 2-6).
4. Extend the lower part of the excitation curve in a straight line until it intersects line A-B drawn in step
3 (see line C-D in Figure 2-6).
The curve (CPB) formed by these two lines now represents the modified excitation characteristics as a
function of the peak voltages that can be produced. After the curve has been drawn, calculate the
following corresponding excitation current I.
V
S
= 2 2 (V
DIFF
) (EQUATION 8)
V
S
= voltage coordinate for determining I
V
DIFF
= differential voltage setting of the BE1-87B
NOTE
The first term in Equation 6 reduces the nI if all the CTs have the same
characteristics. The second term in Equation 6 represents the current (I
R
)
drawn by the relay just at the operating point. It can be calculated as follows:
5000
) (V (2)
I
DIFF
R
=
(EQUATION 9)
Sample Calculation
The various steps for determining the setting of the BE1-87B in a typical bus application will be
demonstrated with the aid of a worked example. Assume the protected zone includes five breakers, all
rated at 69 kV, 1500 MVA, and 1200 amperes, with a maximum interrupting rating of 12,500 amperes.
The excitation curve for the 1200/5 bushing CTs in these breakers is shown in Figure 2-6.
A current sensitivity setting of 0.5 amperes will be used. The voltage tap setting will be determined by
using Method 1 described in the preceding paragraphs. The value of R
S
from Figure 2-6 is (0.0019) (240)
+ 2 (0.0347) = 0.525
Ω. It is assumed that this resistance corresponds to the maximum expected
operating temperature. It is further assumed that the longest CT cable run is 442 feet, and number 10
AWG copper wire is used. The one-way cable resistance at 25 degrees C is 0.450
Ω. The resistance
value of the wire at 25 degrees C or at any temperature T1 may be corrected to any temperature T2 by
means of the following equation.
RT2 = [1 +
ρ1
(T2-T1)] RT1
RT2 = Resistance in ohms at T2, degrees C
RT1 = Resistance in ohms at T1, degrees C
ρ1
= Temperature coefficient of resistance at T1
For standard annealed copper,
ρ1
= 0.00385 at T1 = 25 degrees C. If the maximum expected operating
temperature is assumed to be 50 degrees C, then the following applies.
9282300990 Rev P BE1-87B Application 2-15